The following three constructions have something in common:

**Kernels:**If and are two group homomorphisms, then the composite is the trivial homomorphism if and only if the image of is contained in the kernel of .**Polynomial rings:**If is any -algebra, then an -algebra homomorphism is entirely determined by where it sends .**Topological products:**The product topology on a product of topological spaces is a little counterintuitive—at first glance, the box topology seems more natural—but the product topology has the property that a function is continuous if and only if each component map is continuous.

Do you already see the similarity? Let me rephrase them to make it more obvious:

**Kernels:**If is a group homomorphism, and is any group, then group homomorphisms are in canonical one-to-one correspondence with homomorphisms whose composite with is trivial.**Polynomial rings:**If is any -algebra, then -algebra homomorphisms are in canonical one-to-one correspondence with elements of .**Topological products:**If and are any topological spaces, then continuous functions (with the product topology) are in canonical one-to-one correspondence with tuples of continuous functions , , etc.

These are *universal properties*, and they all have the following rough format:

Given some sort of data one could attach to an arbitrary object …

(e.g. group homomorphisms whose composite with is zero, or elements of ‘s underlying set, or a tuple of continuous functions ,)

…this type of data comes in one-to-one correspondence with maps from to some “universal object,” or the other way around.

(e.g. group homomorphisms from to the , or algebra homomorphisms from to , or continuous functions from to the product space .)

In principle, that’s all there is to an object with a universal property: it’s some object such that maps to it from (or from it to) some arbitrary object correspond to some kind of data you could attach to .

But the reason this notion is so powerful is that having a universal object for some kind of data—let’s say that the set of all possible data on is denoted , and that the universal object is —tells us three related things about and :

**varies “functorially” with .**In other words, if we have a map , we should either consistently get a function or one .

In the kernels example, we have the “data” function given by

Given a group homomorphism , and a piece of data on , we get a composite homomorphism , whose composite with is trivial if is. So we get a function , a way to “pull back” data on to data on .

We also get a notion of pullback in the case of topological products, for which the data on a space is a tuple of maps out of . If is a tuple of continuous functions, and is any continuous map, then the tuple of composites is again a tuple of continuous functions, so we again can “pull back” data from to .

In both of these cases, from a map we get a map in the “other direction,” but in the case of polynomial rings, the relevant notion of “data” on an algebra is just an element of , so as sets. Then from an algebra homomorphism , we just forget the ring structure and get a function , letting us “push forward” elements of to elements of . When data can be “pushed forward” along maps , we say that the data function is a *covariant functor*; if such a map gives a “pullback” instead, we say that the functor is *contravariant*.

Question: The two

contravariantdata functors we’ve considered are the ones for which the maps gotothe universal object (the kernel and product), and thecovariantdata functor is the one where the maps gofromthe universal object (the polynomial ring). Does it always work out like that?

Answer: Yes! If maps going from *to* the universal object correspond to pieces of data on , like maps to the kernel or maps to the product, then we have a collection of bijections , the set of maps from to . This latter set varies contravariantly in : if we have a map , then we get a function sending a map to the composite . This gives a notion of pullback, making it a contravariant functor even if it wasn’t one.

In a similar way, if we have a collection of bijections , as in the case of elements of an -algebra corresponding to -algebra homomorphisms , then the fact that varies *covariantly* with makes a covariant functor.

So if you have a notion of data you can attach to an object, and are wondering whether it might be parameterized by maps to or from some universal object, first see whether the data varies contravariantly or covariantly.

**The universal object comes with a canonical piece of data .**

Continuous functions from a space to the topological product correspond to tuples of maps , , etc., and comes with such a tuple: the projections , , etc.

Given a group homomorphism , group homomorphisms to parameterize maps to whose composite with is trivial. And indeed, itself has a canonical group homomorphism to , the inclusion of it as a subgroup, and the composition of that homomorphism with is trivial.

And for the case where the data on an -algebra is an element of , the universal object has a distinguished element: .

In general, if is the notion of data and is the universal object, we get a distinguished element of . This always happens because, whether the data comes with bijections or , we always get a bijection by considering the case . Since has a distinguished element, the identity map , we get a corresponding distinguished element of .

You can check that this is where the distinguished data on our universal objects came from. This amounts to the following trivial facts:

- The canonical projections are the ones that zip together into the identity map .
- The element is the image of under the identity map .
- The inclusion of into is the composite of the inclusion with the identity map on .

Yes, the canonical piece of data on the universal object is usually not very interesting in itself. But in fact, every piece of data on every object can be obtained from it!

**Every piece of data is either the pullback or pushforward**(according to whether is contravariant or covariant)**of the universal piece of data , and in a unique way.**

Indeed, let’s say is contravariant, and we have some piece of data for some object . Then corresponds to some map by the universal property. What happens when we pull back along ? In terms of hom-sets, the pullback map is just the precomposition map , so the image of in is the element corresponding to the composite , namely the element .

In the case of products, this tells you that every tuple of maps arises as the universal tuple precomposed with a unique map . In the case of polynomial rings, it tells you that for any element of an -algebra , there’s a unique -algebra homomorphism sending to .

This is how mathematicians tend to think about universal objects: they are universally *equipped with some data*, in that every other object with that data gets its data via pullback or pushforward from an appropriate map to the universal object:

- The topological product is the
*universal topological space equipped with continuous maps to each*. - The kernel of is the
*universal group equipped with a group homomorphism to whose composite with is trivial.* - The -algebra is the
*universal -algebra equipped with a distinguished element*.

Question: You keep saying “the” universal object with such-and-such property. How do I know there’s only one? Might the box topology on have the same universal property as the product topology?

Answer: One of the strengths of universal objects is that, if they exist, *they are unique up to unique isomorphism*. So in particular, whenever the box topology on is different from the product topology, it won’t have the property that a function to it is continuous if and only if each component is continuous.

Here’s the proof that any two objects and with the same universal property are isomorphic. Say maps to both and parameterize the contravariant functor —in particular, we are assuming that already comes with a notion of pulling back pieces of data. Then the distinguished element gives a map such that is the pullback of . Similarly, gives a map making the pullback of . Then the composite is a map pulling back to itself; it must therefore equal the identity map which has that same property. Similarly, must equal . Therefore and are inverses of each other and .

So any two objects with the same universal property are isomorphic; moreover, that isomorphism is unique if we require it to pull back or push forward the universal pieces of data to each other. (For all I know, even if the box and product topologies on a product are different, the resulting topological spaces might be abstractly isomorphic. But there won’t be an isomorphism between them that preserves each projection , because the only function that does is the identity.)

Question: So there’s always at most one object with a given universal property, if we count isomorphic objects as “the same.” Is there always

at leastone universal object?

Answer: No, not every universal property is manifested by some universal object. An important case is that of fields: There is no *universal field extension of equipped with a distinguished element* analogous to for -algebras (which don’t need to be fields). The obvious guess would be the fraction field , but the image of under a field homomorphism can never hit zero, since also has to go somewhere. The same holds for every nonzero polynomial in , so can never be sent to an algebraic element over . However, that’s the only problem, so is actually the universal field extension of equipped with a distinguished *transcendental* element!

We can identify other field extensions as having similar universal properties: is the *universal field extension of equipped with a square root of *, for example, since for any choice of field homomorphism and choice of element whose square is , there is a unique field homomorphism sending to . In general, for any irreducible polynomial over a field , the field extension is the *universal field extension of equipped with a root of *.

Question: Does every mathematical object have a universal property?

Answer: Sometimes people do wonder whether a certain object they’ve discovered can be characterized in terms of a universal property. But this always means “in terms of a *nice* universal property,” because every object has a kind of self-referential universal property: we could just define the data on an object to be a map from to . Then is universal with that property, and the distinguished piece of data has is the identity map . Not very interesting!

I wonder if this has applications to physics. For example, an electric field e(x,y,z) can be considered as a set of data that exists at all points (x,y,z) in space. However, if we are dealing with electric fields that are produced by sets of point charges, the finite list of point charges and their positions implies all the e(x,y,z) data. If the inverse problem of finding the point charges from the e(x,y,z) data can be solved then the list of point charges and their positions is analogous to a universal object for such electric fields.

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Thanks for the comment, Stephen! Finding a universal object is a lot like solving an inverse problem, or solving one approximately. If you had an electric field and wanted to model it as one generated by, say, three point charges, we could try to solve for the three point charge locations whose combined electric field best approximated the one we started with. Then that really is an example of a universal property, where the category has arrangements of the point charges as objects, and there’s a morphism from one arrangement to another if the first one is better approximation to the electric field than the second one. The best approximation is then universal among all approximations.

Any best/worst/greatest/least problem works the same way, and every universal property has a similar feel. Given any functorial data you can attach to objects in a category, those data themselves form a category, where the morphisms are ways of pulling back (or pushing forward) one datum to another. The universal datum, then, is one with a unique morphism to every other, just like the best approximation has one morphism to every worse one.

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